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# Express in terms of sine and cosine Calculator

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###  Difficult Problems

1

Here, we show you a step-by-step solved example of express in terms of sine and cosine. This solution was automatically generated by our smart calculator:

$\frac{1-\tan\left(x\right)}{1+\tan\left(x\right)}$
2

Multiply and divide the fraction $\frac{1-\tan\left(x\right)}{1+\tan\left(x\right)}$ by the conjugate of it's denominator $1+\tan\left(x\right)$

$\frac{1-\tan\left(x\right)}{1+\tan\left(x\right)}\frac{1-\tan\left(x\right)}{1-\tan\left(x\right)}$
3

Multiplying fractions $\frac{1-\tan\left(x\right)}{1+\tan\left(x\right)} \times \frac{1-\tan\left(x\right)}{1-\tan\left(x\right)}$

$\frac{\left(1-\tan\left(x\right)\right)\left(1-\tan\left(x\right)\right)}{\left(1+\tan\left(x\right)\right)\left(1-\tan\left(x\right)\right)}$
4

When multiplying two powers that have the same base ($1-\tan\left(x\right)$), you can add the exponents

$\frac{\left(1-\tan\left(x\right)\right)^2}{\left(1+\tan\left(x\right)\right)\left(1-\tan\left(x\right)\right)}$

The first term ($a$) is $1$.

The second term ($b$) is $\tan\left(x\right)$.

The sum of two terms multiplied by their difference is equal to the square of the first term minus the square of the second term. In other words: $(a+b)(a-b)=a^2-b^2$.

$\frac{\left(1-\tan\left(x\right)\right)^2}{1^2-\tan\left(x\right)^2}$

Calculate the power $1^2$

$\frac{\left(1-\tan\left(x\right)\right)^2}{1-\tan\left(x\right)^2}$
5

The sum of two terms multiplied by their difference is equal to the square of the first term minus the square of the second term. In other words: $(a+b)(a-b)=a^2-b^2$.

$\frac{\left(1-\tan\left(x\right)\right)^2}{1-\tan\left(x\right)^2}$

Square of the first term: $\left(1\right)^2 = . Double product of the first by the second:$2\left(1\right)\left(-\tan\left(x\right)\right) = .

Square of the second term: $\left(-\tan\left(x\right)\right)^2 = A binomial squared (difference) is equal to the square of the first term, minus the double product of the first by the second, plus the square of the second term. In other words:$(a-b)^2=a^2-2ab+b^2\frac{1^2+2\cdot 1\cdot -\tan\left(x\right)+\left(-\tan\left(x\right)\right)^2}{1-\tan\left(x\right)^2}$Any expression multiplied by$1$is equal to itself$\frac{1^2+2\cdot -\tan\left(x\right)+\left(-\tan\left(x\right)\right)^2}{1-\tan\left(x\right)^2}$Multiply$2$times$-1\frac{1^2-2\tan\left(x\right)+\left(-\tan\left(x\right)\right)^2}{1-\tan\left(x\right)^2}$Calculate the power$1^2\frac{1-2\tan\left(x\right)+\left(-\tan\left(x\right)\right)^2}{1-\tan\left(x\right)^2}$Simplify$\left(-\tan\left(x\right)\right)^2\frac{1-2\tan\left(x\right)+\tan\left(x\right)^2}{1-\tan\left(x\right)^2}$6 A binomial squared (difference) is equal to the square of the first term, minus the double product of the first by the second, plus the square of the second term. In other words:$(a-b)^2=a^2-2ab+b^2\frac{1-2\tan\left(x\right)+\tan\left(x\right)^2}{1-\tan\left(x\right)^2}$7 Applying the trigonometric identity:$1+\tan\left(\theta \right)^2 = \sec\left(\theta \right)^2\frac{\sec\left(x\right)^2-2\tan\left(x\right)}{1-\tan\left(x\right)^2}$8 Applying the secant identity:$\displaystyle\sec\left(\theta\right)=\frac{1}{\cos\left(\theta\right)}\frac{\frac{1}{\cos\left(x\right)^2}-2\tan\left(x\right)}{1-\tan\left(x\right)^2}$9 Applying the tangent identity:$\displaystyle\tan\left(\theta\right)=\frac{\sin\left(\theta\right)}{\cos\left(\theta\right)}\frac{\frac{1}{\cos\left(x\right)^2}+\frac{-2\sin\left(x\right)}{\cos\left(x\right)}}{1-\tan\left(x\right)^2}$10 The least common multiple (LCM) of a sum of algebraic fractions consists of the product of the common factors with the greatest exponent, and the uncommon factors$L.C.M.=\cos\left(x\right)^2$11 Obtained the least common multiple (LCM), we place it as the denominator of each fraction, and in the numerator of each fraction we add the factors that we need to complete$\frac{1}{\cos\left(x\right)^2}+\frac{-2\sin\left(x\right)\cos\left(x\right)}{\cos\left(x\right)^2}$12 Combine and simplify all terms in the same fraction with common denominator$\cos\left(x\right)^2\frac{\frac{1-2\sin\left(x\right)\cos\left(x\right)}{\cos\left(x\right)^2}}{1-\tan\left(x\right)^2}$Rewrite$1-\tan\left(x\right)^2$in terms of sine and cosine functions$1-\tan\left(x\right)^2$Apply the trigonometric identity:$\tan\left(\theta \right)^n=\frac{\sin\left(\theta \right)^n}{\cos\left(\theta \right)^n}$, where$n=21+\frac{-\sin\left(x\right)^2}{\cos\left(x\right)^2}$Combine all terms into a single fraction with$\cos\left(x\right)^2$as common denominator$\frac{\cos\left(x\right)^2-\sin\left(x\right)^2}{\cos\left(x\right)^2}$Applying the trigonometric identity:$\cos\left(\theta \right)^2-\sin\left(\theta \right)^2 = \cos\left(2\theta \right)\frac{\cos\left(2x\right)}{\cos\left(x\right)^2}$In the original expression, replace the$1-\tan\left(x\right)^2$with$\frac{\cos\left(2x\right)}{\cos\left(x\right)^2}\frac{\frac{1-2\sin\left(x\right)\cos\left(x\right)}{\cos\left(x\right)^2}}{\frac{\cos\left(2x\right)}{\cos\left(x\right)^2}}$13 Rewrite$1-\tan\left(x\right)^2$in terms of sine and cosine functions$\frac{\frac{1-2\sin\left(x\right)\cos\left(x\right)}{\cos\left(x\right)^2}}{\frac{\cos\left(2x\right)}{\cos\left(x\right)^2}}$We can simplify the quotient of fractions$\frac{\frac{1-2\sin\left(x\right)\cos\left(x\right)}{\cos\left(x\right)^2}}{\frac{\cos\left(2x\right)}{\cos\left(x\right)^2}}$by inverting the second fraction and multiply both fractions$\frac{\left(1-2\sin\left(x\right)\cos\left(x\right)\right)\cos\left(x\right)^2}{\cos\left(x\right)^2\cos\left(2x\right)}$Simplify the fraction$\frac{\left(1-2\sin\left(x\right)\cos\left(x\right)\right)\cos\left(x\right)^2}{\cos\left(x\right)^2\cos\left(2x\right)}$by$\cos\left(x\right)^2\frac{1-2\sin\left(x\right)\cos\left(x\right)}{\cos\left(2x\right)}$14 Simplify the fraction$\frac{\frac{1-2\sin\left(x\right)\cos\left(x\right)}{\cos\left(x\right)^2}}{\frac{\cos\left(2x\right)}{\cos\left(x\right)^2}}\frac{1-2\sin\left(x\right)\cos\left(x\right)}{\cos\left(2x\right)}$##  Final answer to the problem$\frac{1-2\sin\left(x\right)\cos\left(x\right)}{\cos\left(2x\right)}\$

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